first, if i wanted to find the x-component of the E-field at point P..

$$E=\int{\frac{k(λdx)}{r^2}}cosθ$$

$$r=sqrt{z^2+x^2}$$

cosθ=$$\frac{x}{sqrt{z^2+x^2}}$$

$$E=kλ\int{\frac{1}{z^2+x^2}}\frac{x}{sqrt{z^2+x^2}}dx$$

evaluated from 0 to L yields..

$$-kλ(\frac{1}{z^2+L^12}-\frac{1}{sqrt{z^2}})$$

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when we did found the potential at piont P in class, we did the following though

$$V_{p}= -\int{\vect{E}\cdot d\vect{r}cosθ$$

my confusion is, when we add the cosθ term, aren’t we just finding the voltage of the x-component E-field?? Because, i thought we were trying to find the total voltage at point P, but this suggests otherwise doesn’t it?

also..

this one’s more of a math question,

expanding $$V_{p}$$..

since $$r=sqrt{z^2+x^2}$$

$$dr=\frac{1}{2}(z^2+x^2)^{-1/2}(2xdx)$$

$$V_{p}= \int{\frac{k(λdx)}{z^2+x^2}\frac{x}{sqrt{z^2+x^2}}(\frac{1}{2}(z^2+x^2) ^{-1/2}(2xdx))=kλ\int{\frac{x^2}{(z^2+x^2)^2}}dx$$

(evaluated from 0 to L)

how do i integrate this?

http://ift.tt/1fumfXc